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• In both, the events meiosis cell division occurs.
• Microsporogenesis results in the formation of microspores while megasporogenesis forms megaspores.

Germ pore is the exine lacking region of pollen grains through which the pollen tube or germ tube emerges soon after pollination.

Exine is the outermost layer of pollen grain which is made up of a highly resistant organic material called sporopollenin. Exine should be hard to withstand high temperature, strong acids and alkali. Germ pores are prominent apertures where sporopollenin is absent which later on protrudes out as pollen tube.

Balanced chemical equation for the reaction is 

4NO(g) + 6H₂O(g) ⇌ 4NH₃(g) + 5O₂(g)

(a) [Ag⁺]² [\(CrO_4^{2-}\)] 

Democritus termed the smallest particles of matter as atoms.

The shell-wise distribution of electrons is called the electronic configuration.

The outermost shell of fluorine atom is L.

Atomic number: 2, Atomic mass number: 5.

Electronic configuration of element is 2, 1 and the element is Lithium. 

The element will get nearest inert gas Helium configuration by loosing one electron. So its valency is 1.

If the valency of an element is 3 then there is a possibility of 3 or 5 electrons in the outermost orbit.

Correct option is (c) oxygen

Distance of a point P(a,b, c) on x-axis

√(b² + c²) 

According to the question,

Number of mice =18,

Number of groups = 3

Since the groups are equally large,

The number of mice in each group can be = 6 mice

The number of ways of placement of mice =18!

For each group the placement of mice = 6!

Hence, the required number of ways = 18!/(6!6!6!)

= 18!/(6!)³

Given f : R→R and g : R→R defined by
f (x) = sin x and g(x) = 5x²
gof(x) = g [f(x)] = g (sin x) = 5 (sin x)² = 5 sin² x

Here, vector a = 2i - j + 2k

and vector b = -i + j - k

∴ vector(a + b) = i + 0, j + k = i + k

∴ vector|a + b| = √(1² + 0² + 1²) = √2

So, unit vector in the direction of 

vector(a + b) = vector((a + b)/|a + b|) = (i + k)/√2

= (1/√2)i + (1/2)j

The correct answer is A

The correct answer is B

The correct answer is(C) 

(i) 1.3 × 10

1.3 × 10 = 13.0

(ii) 36.8 × 10

36.8 × 10 = 368.0

The correct answer is (C)

cost price of 27.5 m @ Rs 53.50 is Rs 1471.25

(b) 0.06
\(\frac{489.1375 \times0.0483\times1.956}{0.0873\times92.581\times 99.749}\) = \(\frac{489\times0.05\times2}{0.09\times93\times100}\)

= \(\frac{489}{9\times93\times10}\) = \(\frac{163}{279}\times\frac1{10}\) = \(\frac{0.58}{10}\) = 0.058 = 0.06.

The correct answer is (B) 

The correct answer is 2.435Km, 2 87/200Km

The correct answer is (C)

(c) 200

In the beginning, the three classes had 2x, 3x and 5x students where x is a constant of proportionality. 40 students were added in each section.

⇒ There are 2k + 40, 3k + 40 and 5k + 40 number of students in each section.

Given, 2k + 40 : 3k + 40 : 5k + 40 = 4 : 5 : 7

⇒ \(\frac{2k+40}{3k+40}=\frac45\) ⇒ 10k + 200 = 12k + 160

⇒ 2k = 40 ⇒ k = 20

∴ Originally the total number of students was 2 × 20 + 3 × 20 + 5 × 20 = 200.

The correct answer is (C)

The correct answer is False

The correct answer is (A)

(c) 16 : 65

Let the original quantity of dettol in the bottle be x litres. 

Then, quantity of water in the bottle = 0 litres 

After the 1st operation : 

Quantity of dettol in the bottle = \(\big(x-\frac{x}{3}\big)\) litres = \(\frac{2x}{3}\) litres

Quantity of water in the bottle = \(\frac{x}{3}\) litres

After the 2nd operation : 

Quantity of dettol in the bottle = \(\big(\frac{2x}{3}-\frac13\times\frac{2x}{3}\big)\) litres = \(\frac{4x}{9}\) litres

∴ Quantity of water in the bottle = \(\big(\frac{x}{3}+\frac{2x}{9}\big)\) litres

After the third operation : 

Quantity of dettol in the bottle  = \(\big(\frac{4x}{9}-\frac13\times\frac{4x}{9}\big)\) litres = \(\frac{8x}{27}\) litres

∴ Quantity of water in the bottle = \(\big(\frac{x}{3}+\frac{2x}{9}+\frac{4x}{27}\big)\) litres

After the fourth operation : 

Quantity of dettol in the bottle  = \(\big(\frac{8x}{27}-\frac13\times\frac{8x}{27}\big)\) litres = \(\frac{16x}{81}\) litres

Quantity of water in the bottle = \(\big(\frac{x}{3}+\frac{2x}{9}+\frac{4x}{27}+\frac{8x}{81}\big)\) litres 

= \(\big(\frac{27x+18x+12x+8x}{81}\big) \) litres

 = \(\frac{65x}{81}\) litres

∴ Required ratio = \(\frac{16x}{81}\): \(\frac{65x}{81}\) = 16 : 65

(d) 5 : 2

Let the required ratio be x : y. 

Milk in x litres of 1st mixture = \(\big(x\times\frac49\big)\) litres = \(\frac{4x}{9}\) litres

Water in x litres of 1st mixture = \(\big(x-\frac{4x}9\big)\) llitres = \(\frac{5x}{9}\) litres

Milk in y litres of 2nd mixture =  \(\big(y\times\frac56\big)\) llitres = \(\frac{5y}{6}\) litres

Water in y litres of 2nd mixture =  \(\big(y-\frac{5y}6\big)\) llitres = \(\frac{y}{6}\) litres

∴ Milk : Water = \(\big(\frac{4x}{9}+\frac{5y}{6}\big)\) : \(\big(\frac{5x}{9}+\frac{y}{6}\big)\)

= \(\big(\frac{8x+15y}{18}\big):\big(\frac{10x+3y}{18}\big)\)

=  (8x + 15y) : (10x +3y)

Given, \(\frac{8x+15y}{10x+3y}=\frac54\) ⇒ 32x + 60y = 50x + 15y

⇒ 18x = 45y ⇒ \(\frac{x}{y} = \frac{45}{18}=\frac52\)

∴ Required ratio = 5 : 2

The reciprocal of 4/7 is 4/7

False

1 is the only number which is its own reciprocal. 

True

2/3 of 8 is same as 2/3 ÷ 8.

False

(c) 53 : 115

Ratio of milk and water

= \(\big(\frac16\times5+\frac38\times4+\frac{5}{12}\times5\big)\) : \(\big(\frac56\times5+\frac58\times4+\frac{7}{12}\times5\big)\)

= \(\big(\frac56+\frac32+\frac{25}{12}\big)\) : \(\big(\frac{25}6+\frac52+\frac{35}{12}\big)\)

= \(\big(\frac{10+18+25}{12}\big)\): \(\big(\frac{50+30+35}{12}\big)\)

= \(\frac{53}{12}:\frac{115}{12}\) = 53 : 115

24% of 150 = 24/100 x 150 = 36

The remaining of that number 

= 150 – 36 = 114

The  answer is False

The correct answer is (B)

(a) 122 : 58 = 61 : 29

Let the quantity of mixture in each container be x. 

Then, Sulphuric acid in 1st container = \(\frac{3x}{5}\)

Water in 1st container = \(\frac{2x}{5}\)

Sulphuric acid in 2nd container = \(\frac{7x}{10}\)

Water is 2nd container = \(\frac{3x}{10}\)

Sulphuric acid in 3rd container = \(\frac{11x}{15}\) 

Water in 3rd container = \(\frac{4x}{15}\)

∴ Required ratio

= \(\bigg(\frac{3x}{5}+\frac{7x}{10}+\frac{11x}{15}\bigg):\) \(\bigg(\frac{2x}{5}+\frac{3x}{10}+\frac{4x}{15}\bigg)\)

= \(\bigg(\frac{36+42+44}{60}\bigg):\)\(\bigg(\frac{24+18+16}{60}\bigg)\)

= 122 : 58 = 61 : 29.

62% = 62/100 = 31/50 = (Fraction) 

= 0.62 (decimal form) 

= 31:50 (ratio)

The correct answer is False

(b) \(\frac13\)

Let the barrel contains wine 3x litres and water x litres. 

Then, total mixture = 4x litres 

Let the part of mixture drawn out be p litres.

∴ (4\(x\) - p) x \(\frac34:(4x-p)\times\frac14+p=1:1\)

⇒ \(3x-\frac{3p}{4}:x-\frac{p}{4}+p=1:1\)

⇒ \(3x-\frac{3p}{4}=x+\frac{3p}{4}⇒2x=\frac{6p}{4}\)

⇒ p = \(\frac{2x\times4}{6}=\frac13(4x)\)

∴ \(\frac13\) part of mixture is drawn out.

The correct answer is False

The correct answer is True

The correct answer is True

i. Newlands was next after Dobereiner who attempted to classify elements according to their properties. 

ii. Newlands arranged all the 56 existing elements at that time in an increasing order of their atomic masses. 

iii. He found that every eighth element had properties similar to that of the first as observed in musical octaves. 

iv. For example:

In the above example, properties of elements belonging to each vertical group are similar. In the first group, properties of H, F and Cl are similar. In the second group, properties of Li, Na and K are similar and so on. 

v. However, Newlands’ octave law was not successful in classifying all the discovered elements. 

vi. After calcium, every eighth element did not possess properties similar to that of the first in the octaves.

Correct answer is (A) first

Three important objects of sculpture made by Harappans are 17.5 cm long statue of limestone of a yogi, bronze statue of a dancing girl and models of carts.